Discrete orderings on $\mathbb{Z}[x,y]$ that violate the universal theory of the integers Working in the language of ordered rings, which we take to have type $(+ - \times < 0\, 1)$, can anyone give an example of a discrete ordering on the polynomial ring in two variables $\mathbb{Z}[x,y]$ such that the resulting ordered ring does not satisfy the universal theory of the integers? 
It is not difficult to show that as a ring, i.e. forgetting the less-than symbol, the ring $\mathbb{Z}[x,y]$ does in fact satisfy the universal theory of the ring of integers. The problem is to find a discrete ordering on $\mathbb{Z}[x,y]$ such that some system of inequalities is solvable in $\mathbb{Z}[x,y]$ but not in the integers, or, on the contrary, to prove that there is no such ordering. 
 A: Here is an example of a discrete ordering on $ \mathbb{Z}[x,y]$ which does not satisfy the universal theory of the ordered ring $ \mathbb{Z}$. 
Let $\alpha$ be a real algebraic number of degree at least 3 over $\mathbb{Q}$. Let $e$ be any irrational between 1 and 2. Let $A$ be the ring 
$\mathbb{Z}[x,\alpha x+x^{-e}]$.  For $f,g\in A$ define $f<g$ to mean that, regarding $f$ and $g$ as functions on the positive reals, $f(r)<g(r)$ for all sufficiently positive $r$. This makes $A$ into an ordered ring. I claim that


*

*$A$ is ring-isomorphic to $\mathbb{Z}[x,y]$. 

*$A$ fails to satisfy the universal theory of the integers.

*$A$ is discretely ordered.
Proof of 1. We show that if $p(x,y)\in \mathbb{Z}[x,y]$ and $p(x,\alpha x+x^{-e})=0$, then $p$ is the zero-polynomial. 
Write $p$ in the form
$$\sum_n q_n(x)x^{-en},$$
where the $q_n$ are polynomials in $\mathbb{Q}(\alpha)[x]$. It follows from the irrationality of $e$ that no monomial in $q_n(x)x^{-en}$ can appear in $q_m(x)x^{-em}$ if $n\ne m$. Therefore all of the $q_n$ must be zero-polynomials, whence p=0. 
Proof of 2. Fix positive integers $a$ and $b$ such that $1<a/b<\epsilon$. Roth's theorem on diophantine approximation (here) implies that there is a natural number $N$ such that the following holds for all integers $x$ and $y$: 
$$x>N\implies|\alpha x-y|^bx^a>1.$$
Using quantifier elimination for real closed fields and the assumption that $\alpha$ is  algebraic,  it is possible to construct a quantifier-free formula, call it $\phi(x,y)$, that  is equivalent  in any real closed field to the above inequality. (In an arbitrary real closed field $\alpha$ has the meaning of an element satisfying a certain polynomial and lying in a certain interval with rational endpoints.)
Then  $$\mathbb{Z}\models \forall x,y\,\phi(x,y).$$ On the other hand, in the real closure of the ring $A$, with $u=x$ and $v=\alpha x+x^{-e}$, it is easily verified (using $1<a/b<\epsilon$) that 
$$u>N\textrm{ and }|\alpha u-v|^bu^a<1.$$
Since $\phi$ is is quantifier-free, it follows that
$$A\models \neg\phi(u,v).$$
Therefore $A$ fails to satisfy the universal theory of $\mathbb{Z}$.
Proof of 3. We will show that A has no finite elements other than the integers. It is enough to show that if  $H\in \mathbb{Z}[x,y]$ is non-constant then    $H(x,\alpha x+x^{-e})$ has at least one monomial of positive degree. We shall do this first for homogeneous $H$ and then for sums of non-constant homogeneous polynomials of different degrees.
Assume now that $H(x,y)$ is homogeneous of degree $n>0$. Let $$h(y)=H(1,\alpha+y).$$
Writing out the Taylor expansion for $h$ about $y=0$,
$$h(y)=\sum_{k=0}^{n}\dfrac{h^{(k)}(\alpha)}{k!}y^k,$$
where $h^{(k)}$ is the $k$-th derivative of $h$. Substituting $x^{-1-e}$ for $y$, and multiplying both sides by $x^n$, we obtain
\begin{equation}\tag{$*$} H(x,\alpha x+x^{-e})=\sum_{k=0}^{n}\dfrac{h^{(k)}(\alpha)}{k!}x^{n-k(1+e)}.\end{equation}
The exponent $n-k(1+e)$ is positive precisely when $k=0,1,\ldots,\lfloor n/(1+e)\rfloor$. By way of contradiction, suppose that $h^{(k)}(\alpha)$ vanishes for all of these values of $k$. 
Let $f$ be the irreducible polynomial of $\alpha$ over $\mathbb{Q}$, and let $m$ be the degree of $f$. Then $f^{1+\lfloor n/(1+e)\rfloor{}}$ must divide $h$.  Since $h$ has degree at most $n$, we find that $$\tag{$**$}m(1+\lfloor n/(1+e)\rfloor)\le n.$$ But  $$n/(1+e)<1+\lfloor n/(1+e)\rfloor,$$ hence 
$$\tag{$***$}mn/(1+e)<m(1+\lfloor n/(1+e)\rfloor).$$
Combining the inequalities $(**)$ and $(***)$, we have $$mn/(1+e)<n,$$ hence $m<1+e$. But by assumption $e<2$. Therefore $m<3$.  This contradicts the definition of $\alpha$, which is assumed to have degree $m\ge3$. This concludes the homogeneous case.
Now suppose $H\in \mathbb{Z}[x,y]$ has the form $\sum_k H_k$, where $H_k$ is homogeneous of degree $k>0$. It follows from equation ($*$) that the powers of $x$ in $H_k(x,\alpha x+x^{-e})$ all have the form $a+be$, with $a+b=k$. Since $e$ is irrational, no power of $x$ in $H_k(x,\alpha x+x^{-e})$ can appear in $H_m(x,\alpha x+x^{-e})$ if $k\ne m$. But we have shown that  some positive power of $x$ appears in $H_k(x,\alpha x+x^{-e})$ for every $k>0$. This completes the proof.
